wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let P(a secθ,b tanθ) and Q(a secϕ,b tanϕ), where θ+ϕ=π2, be two points on the hyperbola x2a2y2b2=1.
If (h, k) is the point of the intersection of the normals at P and Q, then k is equal to


A

a2+b2a

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

(a2+b2a)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

a2+b2b

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

(a2+b2b)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D

(a2+b2b)


Firstly, we obtain the slope of normal to x2a2y2b2=1 at (a secθ,b tanθ)
On differentiating w.r.t.x, we get
2xa22yb2×dydx=0dydx=b2a2xy
Slope for normal at the point (a secθ,b tanθ) will be a2b tanθb2a secθ=absinθ
Equation of normal at (a secθ,b tanθ) is
yb tanθ=absinθ(xa secθ)
(a sinθ)x+by=(a2+b2)tanθ
ax+b y cosecθ=(a2+b2)secθ ... (i)
Similarly, equation of normal to x2a2y2b2=1 at (a secϕ,b tanϕ) is ax+b y cosec ϕ=(a2+b2)secϕ ... (ii)
On subtracting Eq. (ii) from Eq. (i), we get
b(cosecθcosecϕ)y=(a2+b2)(secθsecϕ)
y=a2+b2b.secθsecϕcosecθcosecϕ
But secθsecϕcosecθcosecϕ=secθsec(π2θ)cosecθcosec(π2θ)
[ϕ+θ=π2]
=secθcosecθcosecθsecθ=1
Thus, y=(a2+b2b),i.e.k=(a2+b2b)


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Lines and Points
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon